Range and Null Space

Range and Null Space

Definition: Let T:V→WT:V \rightarrow W be a linear transformation. The set of all vectors y∈Wy \in W such that y=T(x)y = T(x) for some x∈Vx \in V is called the range of TT and denoted by R(T)\mathcal{R}(T).

Definition: Let T:V→WT:V \rightarrow W be a linear transformation. The set of all vectors x∈Vx \in V such that T(x)=0T(x) = 0 is called the null space of TT and denoted by N(T)\mathcal{N}(T).

Let T:V→WT:V \rightarrow W be a linear transformation. Let dim(V)=ndim(V) = n and dim(W)=mdim(W) = m. Recall that,

Theorem: dim(R(T))+dim(N(T))=dim(V)dim(\mathcal{R}(T)) + dim(\mathcal{N}(T)) = dim(V)

Proof: Let B={v1,v2,…,vk}B = \{v_1, v_2, \dots, v_k\} be a basis for N(T)\mathcal{N}(T). Extend BB to a basis B′={v1,v2,…,vk,vk+1,…,vn}B' = \{v_1, v_2, \dots, v_k, v_{k+1}, \dots, v_n\} for VV. Then B′B' is a basis for VV and B′′={T(vk+1),T(vk+2),…,T(vn)}B'' = \{T(v_{k+1}), T(v_{k+2}), \dots, T(v_n)\} is a basis for R(T)\mathcal{R}(T). Hence dim(R(T))=n−k=dim(V)−dim(N(T))dim(\mathcal{R}(T)) = n - k = dim(V) - dim(\mathcal{N}(T)) ■\blacksquare

Definition: Let T:V→WT:V \rightarrow W be a linear transformation. The rank of TT is defined as rank(T)=dim(R(T))rank(T) = dim(\mathcal{R}(T)).

Remark: rank(T)rank(T) is equal to the number of linearly independent columns of TT.

Remark: In general, a basis for R(T)\mathcal{R}(T) and N(T)\mathcal{N}(T) can be found by finding a basis for the column space and null space of the matrix representation of TT with respect to some basis for VV and WW.

Eample: V=R2x2V = \mathbb{R}^{2x2}, W=VW = V, Find a basis for R(A)\mathcal{R}(\mathcal{A})

B={[1000],[0100],[0010],[0001]}C={[1000],[1100],[1110],[1111]}A(x)=Sx+xS⊤=[01−10]x+x[01−10]⊤A=[111−10000−11110−1−10]\begin{align*} B &= \begin{Bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \end{Bmatrix} \\ C &= \begin{Bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \end{Bmatrix} \\ \mathcal{A}(x) &= Sx + x S^\top = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}x + x\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}^\top \\ A &= \begin{bmatrix} 1 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ -1 & 1 & 1 & 1 \\ 0 & -1 & -1 & 0 \end{bmatrix} \\ \end{align*}

Solution: Start with finding a basis for R(A)\mathcal{R}(\mathcal{A})

R(A):={w∈W ∣  ∃v∈V, Av=w}R(A):={y∈R4 ∣  ∃x∈R4, Ax=y}\begin{align*} \mathcal{R}(\mathcal{A}) &:= \{ w \in W \ | \ \ \exist v \in V , \ \mathcal{A}v = w \} \\ \mathcal{R}(A) &:= \{ y \in \mathbb{R}^4 \ | \ \ \exist x \in \mathbb{R}^4 , \ Ax = y \} \\ \end{align*}

Finding a basis for R(A)\mathcal{R}(A) is equivalent to finding a basis for the column space of AA. The basis for the column space of AA can be found by reducing AA to independent columns by elementary column operations and then taking the non-zero columns of the reduced matrix.

A=[111−10000−11110−1−10]∼[10000000−12000−100]\begin{align*} A &= \begin{bmatrix} 1 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ -1 & 1 & 1 & 1 \\ 0 & -1 & -1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 2 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{bmatrix} \\ \end{align*}
basis for R(A)={[10−10],[002−1]}\begin{align*} \text{basis for } \mathcal{R}(A) &= \begin{Bmatrix}\begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 2 \\ -1 \end{bmatrix}\end{Bmatrix} \\ \end{align*}
[10−10]=[w1]C  ⟹  w1=1c1+0c2+(−1)c3+0c4=c1−c3=[1000]−[1110]=[0−1−10]\begin{align*} \begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \end{bmatrix} = [w_1]_C \implies w_1 &= 1c_1 + 0c_2 + (-1)c_3 + 0c_4 = c_1 - c_3 \\ &= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} \\ \end{align*}
[002−1]=[w2]C  ⟹  w2=0c1+0c2+2c3+(−1)c4=2c3−c4=[2220]−[1111]=[111−1]\begin{align*} \begin{bmatrix} 0 \\ 0 \\ 2 \\ -1 \end{bmatrix} = [w_2]_C \implies w_2 &= 0c_1 + 0c_2 + 2c_3 + (-1)c_4 = 2c_3 - c_4 \\ &= \begin{bmatrix} 2 & 2 \\ 2 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \\ \end{align*}
basis for R(A)={[0−1−10],[111−1]}dim(R(A))=2=rank(A)\begin{align*} \text{basis for } \mathcal{R}(\mathcal{A}) &= \begin{Bmatrix}\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix},\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}\end{Bmatrix} \\ dim(\mathcal{R}(A)) &= 2 = \text{rank}(A) \\ \end{align*}
  1. AA is a 4x44x4 matrix, dim(R(A))=4−dim(N(A))dim(\mathcal{R}(A)) = 4 - dim(\mathcal{N}(A))
  2. dim(N(A))=2dim(\mathcal{N}(A)) = 2 since rank(A)=2rank(A) = 2
  3. dim(R(A))=4−2=2dim(\mathcal{R}(A)) = 4 - 2 = 2
  4. dim(R(A))=2dim(\mathcal{R}(\mathcal{A})) = 2

Now, we need to find a basis for N(A)\mathcal{N}(\mathcal{A}).

N(A):={v∈V ∣  Av=0}N(A):={x∈R4 ∣  Ax=0W}\begin{align*} \mathcal{N}(\mathcal{A}) &:= \{ v \in V \ | \ \ \mathcal{A}v = 0 \} \\ \mathcal{N}(A) &:= \{ x \in \mathbb{R}^4 \ | \ \ Ax = 0_W \} \\ \end{align*}

Finding a basis for N(A)\mathcal{N}(A) is equivalent to finding a basis for the null space of AA. The basis for the null space of AA can be found by reducing AA by elementary row operations and then equating the Aˉx=0\bar Ax =0.

A=[111−10000−11110−1−10]∼[100−1011000000000]=Aˉ\begin{align*} A &= \begin{bmatrix} 1 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 \\ -1 & 1 & 1 & 1 \\ 0 & -1 & -1 & 0 \end{bmatrix} \sim \begin{bmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} = \bar A \\ \end{align*}
Aˉx=0[100−1011000000000][x1x2x3x4]=[0000]x1−x4=0  ⟹  x1=x4x2+x3=0  ⟹  x2=−x3\begin{align*} \bar Ax = 0 \\ \begin{bmatrix} 1 & 0 & 0 & -1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} &= \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} \\ x_1 - x_4 = 0 \implies x_1 &= x_4 \\ x_2 + x_3 = 0 \implies x_2 &= -x_3 \\ \end{align*}
[x1x2x3x4]=[x1x2−x2x1]=x1[1001]+x2[01−10]\begin{align*} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} &= \begin{bmatrix} x_1 \\ x_2 \\ -x_2 \\ x_1 \end{bmatrix} = x_1\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} + x_2\begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \end{bmatrix} \\ \end{align*}
basis for N(A)={[1001],[01−10]}\begin{align*} \text{basis for } \mathcal{N}(A) &= \begin{Bmatrix}\begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \end{bmatrix}\end{Bmatrix} \\ \end{align*}
[1001]=[v1]B  ⟹  v1=1b1+0b2+0b3+1b4=b1+b4=[1000]+[0001]=[1001]\begin{align*} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} = [v_1]_B \implies v_1 &= 1b_1 + 0b_2 + 0b_3 + 1b_4 = b_1 + b_4 \\ &= \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \\ \end{align*}
[01−10]=[v2]B  ⟹  v2=0b1+1b2+(−1)b3+0b4=b2−b3=[0100]−[0010]=[01−10]\begin{align*} \begin{bmatrix} 0 \\ 1 \\ -1 \\ 0 \end{bmatrix} = [v_2]_B \implies v_2 &= 0b_1 + 1b_2 + (-1)b_3 + 0b_4 = b_2 - b_3 \\ &= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \\ \end{align*}
basis for N(A)={[1001],[01−10]}dim(N(A))=2=nullity(A)\begin{align*} \text{basis for } \mathcal{N}(\mathcal{A}) &= \begin{Bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\end{Bmatrix} \\ dim(\mathcal{N}(A)) &= 2 = \text{nullity}(A) \\ \end{align*}

Definition: Let T:V→WT:V \rightarrow W be a linear transformation. The nullity of TT is defined as nullity(T)=dim(N(T))nullity(T) = dim(\mathcal{N}(T)).

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